Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\) thì \(a=b.k;c=d.k\)
Ta có:
\(\dfrac{a}{3.a+b}=\dfrac{b.k}{3.b.k+b}=\dfrac{b.k}{b.\left(3k+1\right)}=\dfrac{k}{3k+1}\left(1\right)\\ \dfrac{c}{3.c+d}=\dfrac{d.k}{3.d.k+d}=\dfrac{d.k}{d.\left(3k+1\right)}=\dfrac{k}{3k+1}\left(2\right)\)
Từ \(\left(1\right);\left(2\right)\Rightarrow\dfrac{a}{3.a+b}=\dfrac{c}{3c+d}\)